Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

A particle of charge $q$ and mass $m$ is moving with a constant velocity v along $+x$ - axis. It enters a region of constant magnetic field along $-z$ - axis from $x = a$ to $x = b$. The minimum value of $ v$ required so that the particle can enter the region $x>b$ is

$\begin {array} {1 1} (a)\;\large\frac{qbB}{m} & \quad (b)\;\large\frac{qaB}{m} \\ (c)\;\large\frac{q(b+a)B}{m} & \quad (d)\;\large\frac{q(b-a)B}{m} \end {array}$


Can you answer this question?

1 Answer

0 votes
The particle cannot enter $x>b$ if the radius of the circular path is lesser than the width of the region.
The width of the region is (b-a).
we have v= r B q /m.
For minimum value of v ,r must be equal to (b-a)
Ans : (d)


answered Feb 19, 2014 by thanvigandhi_1
edited Sep 15, 2014 by thagee.vedartham

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App